The geometric sequence $(a_i)$ is defined by the formula: $a_1 = \dfrac{1}{16}$ $a_i = -4a_{i-1}$ What is $a_{3}$, the third term in the sequence?
Answer: From the given formula, we can see that the first term of the sequence is $\dfrac{1}{16}$ and the common ratio is $-4$ To find the third term, we can rewrite the given recurrence as an explicit formula. The general form for a geometric sequence is $a_i = a_1 r^{i - 1}$ . In this case, we have $a_i = \dfrac{1}{16} \left(-4\right)^{i - 1}$ To find $a_{3}$ , we can simply substitute $i = 3$ into the formula. Therefore, the third term is equal to $a_{3} = \dfrac{1}{16} \left(-4\right)^{3 - 1} = 1$.